\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

Có \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)

\(\Rightarrow A< B\)

\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

Do  \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)nên  \(1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

Vậy  \(A< B\)

Ta đi so sánh \(\frac{2017.2018+1}{2017.2018}\)với\(\frac{2018.2019+1}{2018.2019}\)có :

\(\frac{2017.2018+1}{2017.2018}=\frac{2017.2018}{2017.2018}+\frac{1}{2017.2018}=1+\frac{1}{2017.2018}\left(\cdot\right)\)

\(\frac{2018.2019+1}{2018.2019}=\frac{2018.2019}{2018.2019}+\frac{1}{2018.2019}\left(\cdot\cdot\right)\)

\(\frac{1}{2017.2018}>\frac{1}{2018.2019}\left(\cdot\cdot\cdot\right)\)Từ \(\left(\cdot\right);\left(\cdot\cdot\right)\&\left(\cdot\cdot\cdot\right)\Rightarrow\frac{2017.2018+1}{2017.2018}>\frac{2018.2019+1}{2018.2019}\)

\(\Leftrightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}.\)

#)Trả lời :

\(\frac{2017\times2018}{2017\times2018+1}=\frac{0}{1}=0\)

\(\frac{2018\times2019}{2018\times2019+1}=\frac{0}{1}=0\)

\(\Rightarrow\frac{2017\times2018}{2017\times2018+1}=\frac{2018\times2019}{2018\times2019+1}\)

Ta có:

\(\frac{2017.2019}{2018.2018}\)

\(=\frac{2017.\left(2018+1\right)}{\left(2017+1\right).2018}\)

\(=\frac{2017.2018+2017}{2017.2018+2018}\)

Vì \(2017.2018+2017< 2017.2018+2018\)( tử nhỏ hơn mẫu )

\(\Rightarrow\frac{2017.2018+2017}{2017.2018+2018}< 1\)

Vậy \(\frac{2017.2019}{2018.2018}< 1\)

        ( Mk nghĩ vậy )

                          ~~~~~~~Hok tốt~~~~~~~

\(\frac{2017.2019}{2018.2018}=\frac{2017.\left(2018+1\right)}{2018.\left(2017+1\right)}=\frac{2017.2018+2017}{2018.2017+2018}\)

\(2017< 2018\Rightarrow2017.2018+2017< 2018.2017+2018\Rightarrow\frac{2017.2018+2017}{2018.2017+2018}< 1\Rightarrow\frac{2017.2019}{2018.2018}< 1\)

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)

b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2\left(1-\frac{1}{2019}\right)\)

\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)

\(=2.\frac{2018}{2019}\)

\(=\frac{4036}{2019}\)

Phần c tương tự nha

a) \(\frac{1}{1.2}\) +  \(\frac{1}{2.3}\) + .......+  \(\frac{1}{2017.2018}\)

= 1 -  \(\frac{1}{2}\) + \(\frac{1}{2}\) -  \(\frac{1}{3}\) + .......+  \(\frac{1}{2017}\) -   \(\frac{1}{2018}\)

= 1 -  \(\frac{1}{2018}\) =  \(\frac{2017}{2018}\)

câu a) mik sửa đề một tí ko biết có đúng ko

câu b , c tương tự nhưng cần lấy tử ra chung 

a)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)nhóm 2 ra ngoài rồi làm như câu a

c)nhóm 4 ra rồi làm như câu a

cái gì đấy

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Đề là gì vậy bạn !

id nhu 1 tro dua

Ta có 

A= \(\frac{2017^{2018}-3+4}{2017^{2018}-3}=1+\frac{4}{2017^{2018}-3}\)

B= \(1+\frac{4}{2017^{2018}-5}\)

vậy A > B